1. Refraction of Light

2. Light travels in straight lines. However, when light travels through different mediums, its path is deviated. This deviation is called refraction. Each medium has its own level of refraction. We can calculate the index of refraction, n, of a medium using: n = c/v Where c is the speed of an electromagnetic wave in a vacuum, 3.00 x 10 8 m/s and v is the speed of light in that medium in m/s

3. Example Using the table of values below, identify the medium through which light travels when its velocity is 8.57 x 10 7 m/s. n = c/v c= 3.00 x 10 8 m/sn = c/v v= 8.57 x 10 7 m/sn= 3.00 x 10 8 m/s 8.57 x 10 7 m/s n= 3.50, gallium phosphide MediumWater Ethyl alcohol PlexiglassDiamond Gallium phosphide Index of refraction (n) 1.331.361.512.423.50

4. Note: The greater the index of refraction of a medium, the slower the speed at which light will travel through it.

5. A relative index of refraction can also be calculated when a light ray does not pass through a vacuum. The relative index of refraction is a ratio of the index of refractions of the two mediums involved.

6. Example: Calculate the relative index of refraction of a light ray that travels through water and then through Plexiglass. n R = n 2 / n 1 n 1 : index of refraction of the first medium n 2: index of refraction of the second medium n R = 1.51 / 1.33 = 1.14

7. Geometry of Refraction Since light is travelling in two different mediums, the angle of incidence is no longer equal to the angel of refraction. Laws of refraction: 1.The incident ray, the normal and the refracted ray are all on the same plane. n1n1 n2n2 θiθi θRθR Incident light ray refracted light ray

8. 2. The ratio of the sine of the angle of incidence and refraction are equal to the ratio between the index of refraction of the two mediums. sin θ i = n 2 sin θ R n 1 n 1 sin θ i = n 2 sin θ R

9. Example: A light travel from a diamond (n=2.42) into water (n=1.33). If the rays have an angle of incidence of 30 o, calculate the angle of refraction. n 1 sin θ i = n 2 sin θ R 2.42 x sin (30) = 1.33 x sin θ R θ R = 65 o

10. EXAMPLE A light ray travelling through air encounters a glass cup with an angle of incidence of 35 o. The light ray then travels through water and finally through ice. Determine the final angle of refraction once the light ray exits the ice. glassicewater

11. Total Internal Reflection When a light ray travels in a medium with a high refractive index and encounters one with a low refractive index, refraction does NOT always occur; the light ray can be reflected instead. The angle of incidence will determine whether the ray is reflected of refracted.

12. If the angle of incidence is greater than or equal to the critical angle, then internal reflection occurs. If not the ray will be refracted into the second medium. medium 2 medium 1

13. We can determine the value of the critical angle by using θ C = sin -1 (n 2 /n 1 ) Where θ C is the critical angle n 1 is the refractive index of medium 1 n 1 is the refractive index of medium 2

14. Example An aquarium has a built in light source. The light rays hit the aquarium’s glass walls at an angle of incidence of 65 o. a.Determine whether reflection or refraction occurs. b.At what angle will the light ray reflect/refract?